Now construct a decimal number $x = 0.x_1x_2...x_n...$ as follows. We let $x_1 = 0$ if $x_{1,1} = 1$ and we let $x_1 = 1$ if $x_{1,1} = 0$. We let $x_2 = 0$ if $x_{2,2} = 1$ and we let $x_2 = 1$ if $x_{2,2} = 0$. In general, we let $x_n = 0$ if $x_{n,n} = 1$ and we let $x_n = 1$ if $x_{n,n} = 0$. Then $X$ differs from all of the decimal numbers in the list above in at least one decimal spot. So the decimal $x$ corresponds to a subset $X$ of $\mathbb{N}$ that is not on the list above. This is a contradiction since $f$ is assumed to be a bijection.

Therefore the set $\mathcal P(\mathbb{N})$ is uncountable. $\blacksquare$